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I am self-studying https://sayanmuk.github.io/StochasticAnalysisManifolds.pdf, the following proof can be found on the 9th page of the monograph I just provided the reference. It follows from a previous question that I asked:Intuition behind $Q_t=\sum \langle M^{\alpha},M^{\alpha}\rangle_t+\sum |A^{\alpha}|^3_t+|A^{\alpha}|_t+t$

I am having a problem with two steps of the following theorem proof:

$\textbf{Theorem }$ Suppose that $\sigma$ is globally Lipschitz and $X_0$ is square integrable. Then the stochastic differential equation (SDE) $(\sigma, Z, X_0)$ has a unique solution $X = \{X_t, t \geq 0\}$.

Proof: In this proof, we use the notation

$$|Y|_{\infty,t} = \max_{0 \leqslant s \leqslant t} |Y_s|$$

for a vector-valued process $Y$. We solve the equation by the usual Picard's iteration. Define a sequence $\{X^n\}$ of semimartingales by $X^0_{t} = X_0$ and

\begin{equation} X^n_{t} = X^0_{t} + \int_{0}^{t} \sigma(X^{n-1}_s) \, dZ_s \tag{1.1.5} \end{equation}

We claim that $\{X_n\}$ converges to a continuous semimartingale $X$ which satisfies the equation. Define the increasing process $Q$ as in (1.1.3) and let $\eta$ be its inverse as before. Each $\eta_T$ is a stopping time for fixed $T$. By (1.1.4) and the assumption that $\sigma$ is globally Lipschitz, we have

\begin{equation} \mathbb{E} \left[ \sup_{0 \leq s \leq \eta_T} \left|X^n - X^{n-1}\right|^2 \right] \leq C_1 \mathbb{E} \left[ \int_0^{\eta_T} \left| \sigma(X^{n-1,s}) \right|^2 \, ds \right] \end{equation}

\begin{equation} \begin{aligned} \leq C_2 \mathbb{E} \left[ \int_{0}^{\eta_T} \left|X^{n-1,s} - X^{n-2,s}\right|^2 dQ_{\eta_T} \right]. \end{aligned} \end{equation}

Making the change of variable $s = \eta(u)$ in the last integral and using $Q_{\eta(u)} = u$, we have \begin{equation} E \left[ \sup_{0 \leq s \leq \eta_T} \left|X^n - X^{n-1}\right|^2 \right] \leq C_2 E \left[ \int_{0}^{\eta_T} \sup_{0 \leq s \leq \eta_u} \left|X^{n-1}_s - X^{n-2}_s\right|^2 \, du \right]. \end{equation}

For the initial step, from \begin{equation} X^1_{t} - X^0_{t} = \int_{0}^{t} \sigma(X^{0}_s) \, dZ_s = \sigma(Z_0)Z_t. \end{equation}

and (1.1.4) again we have \begin{equation} E \left[ \sup_{0 \leq s \leq \eta_T} \left|X^{1} - X^{0}\right|^2 \right] \leq C_3 E \left[ \left|X_{0}\right|^2 + 1 \right] Q_{\eta_T} = C_3 T E \left[ \left|X_{0}\right|^2 + 1 \right]. \end{equation}

Questions:

  1. I wonder if this is correct:$ X^1_{t} - X^0_{t} = \int_{0}^{t} \sigma(X^{0}_s) \, dZ_s = \sigma(Z_0)Z_t$. As far as I understand this $X^{0}_s$, should be a constant, right? Once it is assumed to be equal to $X_0$. If that is so the integral above would be $\sigma(X_0)Z_t$ instead. So probably, there was a typo. But my problem relies on what comes next:

2)$E \left[ \sup_{0 \leq s \leq \eta_T} \left|X^{1} - X^{0}\right|^2 \right] \leq C_3 E \left[ \left|X_{0}\right|^2 + 1 \right] Q_{\eta_T}$. This step is puzzling me. How does the author get $|X_{0}|^2+1$? Is he using some Lipschitz property of $\sigma$? How does that relate to the previous step, I mentioned in 1). Since he mentions that we just need to apply the inequality $E \max_{0\leqslant t\leqslant\tau}|\int_{0}^{t}F_s dZ_s|^2\leqslant C E \int_0^{\tau}|F_s|^2dQ_s$(see Intuition behind $Q_t=\sum \langle M^{\alpha},M^{\alpha}\rangle_t+\sum |A^{\alpha}|^3_t+|A^{\alpha}|_t+t$ ).

However, if I do apply that inequality I end up with an integral inside expectation of this kind $\int_0^T |\sigma(X_0)|^2 dQ_{\nu_T}$. How does this latter ends up being $(|X|^2+1)Q_{\eta_T}$

I would appreciate it if someone could clarify those last steps.

Thanks in advance.

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1 Answer 1

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  1. I wonder if this is correct:$ X^1_{t} - X^0_{t} = \int_{0}^{t} \sigma(X^{0}_s) \, dZ_s = \sigma(Z_0)Z_t$. As far as I understand this $X^{0}_s$, should be a constant, right? Once it is assumed to be equal to $X_0$. If that is so the integral above would be $\sigma(X_0)Z_t$ instead. So probably, there was a typo. But my problem relies on what comes next:

Yes, but he defined $X_{s}^{0}=X_{0}$, so you are both right.

2)$E \left[ \sup_{0 \leq s \leq \eta_T} \left|X^{1} - X^{0}\right|^2 \right] \leq C_3 E \left[ \left|X_{0}\right|^2 + 1 \right] Q_{\eta_T}$. This step is puzzling me. How does the author get $|X_{0}|^2+1$? Is he using some Lipschitz property of $\sigma$? How does that relate to the previous step, I mentioned in 1). Since he mentions that we just need to apply the inequality $E \max_{0\leqslant t\leqslant\tau}|\int_{0}^{t}F_s dZ_s|^2\leqslant C E \int_0^{\tau}|F_s|^2dQ_s$(see Intuition behind $Q_t=\sum \langle M^{\alpha},M^{\alpha}\rangle_t+\sum |A^{\alpha}|^3_t+|A^{\alpha}|_t+t$ ). However, if I do apply that inequality I end up with an integral inside expectation of this kind $\int_0^T |\sigma(X_0)|^2 dQ_{\nu_T}$. How does this latter ends up being $(|X|^2+1)Q_{\eta_T}$

So indeed they apply the 1.1.4 inequality to bound by

$$\int_0^{\eta_T} |\sigma(X_0)|^2 dQ_{t}$$

and yes they use Lipschitz

$$|\sigma(X_0)|^2 \leq |\sigma(X_0)-\sigma(0)|^2 +|\sigma(0)|^2\leq c_{1} |X_{0}^{2}|+c_{2}\leq c_{3}(|X_{0}^{2}|+1).$$

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